Bifurcation problem Consider a physical system whose state is described by the r
ID: 2984367 • Letter: B
Question
Bifurcation problem
Consider a physical system whose state is described by the real number x. The energy of each possible state is given by E(x) = x2 - beta x2/1 + x2 Here beta is a parameter determined by the environment (ambient temperature and so forth). Draw a bifurcation diagram for all critical points (i.e. points where the derivative vanishes) as beta varies. Draw new bifurcation diagram for only the (global) minima x* of E(x) as beta varies.Explanation / Answer
f = {L, [Psi], d} /. NDSolve[{ d'[t] == (1/ z) ( [Alpha] - [Beta] (d[t]/Y[t]) - [Gamma] L[ t] - [Delta] [Psi][t] - [Tau] Y[t]) + 0.75 (d[t]/Y[t]) d[t], L'[t] == (j (Y[t]/K) - e)/N, [Psi]'[t] == p L[t] - 0.35, Y[t] == (I + E + [Alpha] - [Beta] ( d[t]/Y[t]) - [Gamma] L[ t] - [Delta] [Psi][ t])/(1 - (1 - [Tau]) ([Psi][ t] + (1 - s) (1 - [Psi][t]) - m z)), L[0] == 0.50, [Psi][0] == 0.50, d[0] == 0} /. { [Alpha] -> 0.05, [Gamma] -> 0.75, [Delta] -> 0.75, X -> 1, [Beta] -> 0.05, j -> 0.45, [Tau] -> 0.35, I -> 1, s -> 0.85, p -> 0.75, E -> 1, m -> 1.2, e -> 0.035, z -> 5.5, K -> 3.5, N -> 4}, {L, [Psi], d, Y}, {t, 0, 300}, MaxSteps -> 1000000 ][[1]] l = {L[t], [Psi][t], d[t]}; cps[L] = f /. Quiet[Solve[f[[L'[t]]] == 0], All] Show[GraphicsArray[ Table[Plot[f[[i]][t], {t, 0, 300}, PlotRange -> All, PlotStyle -> Blue, Filling -> 0, AxesLabel -> TraditionalForm /@ {t, l[[i]]}, DisplayFunction -> Identity], {i, 3}] ]] data2 = Table[Evaluate[f[[#]][t] & /@ {1, 2, 3}], {t, 0, 300}]; Export["data2.csv", data2] ListLinePlot[Transpose@data2, Filling -> 0, AxesLabel -> {t, {L, [Psi], d}}, PlotRange -> All] Graphics3D[Point[data2], BoxRatios -> 1, AxesLabel -> TraditionalForm /@ l] ParametricPlot3D[Evaluate[Append[#[t] & /@ f, Red]], {t, 0, 300}, BoxRatios -> {1, 1, 1}, PlotRange -> All, PlotPoints -> 1500, AxesLabel -> TraditionalForm /@ l]
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.